Problem 31 - Calculus for Biology and Medicine

Question

Differentiate $$ g(N)=\frac{b N}{(k+N)^{2}} $$ with respect to \(N\). Assume that \(b\) and \(k\) are positive constants.

Step-by-Step Solution

Answer

The derivative is \(g'(N) = \frac{b(k^2 - N^2)}{(k+N)^4}\).

Step 1: Identify the Derivative Rule

The function \(g(N)\) is a quotient of two functions. Therefore, we should use the quotient rule for differentiation, which states that \(\left[\frac{u}{v}\right]' = \frac{u'v - uv'}{v^2}\), where \(u = bN\) and \(v = (k+N)^2\).

Step 2: Differentiate the Numerator

Differentiate the numerator \(u = bN\) with respect to \(N\). Since \(b\) is a constant, the derivative \(u' = b\).

Step 3: Differentiate the Denominator

Differentiate the denominator \(v = (k+N)^2\) using the chain rule. The derivative is \(v' = 2(k+N)\cdot 1 = 2(k+N)\) since the derivative of \(k+N\) with respect to \(N\) is 1.

Step 4: Apply the Quotient Rule

Substitute \(u'\), \(v\), and \(v'\) back into the quotient rule formula: \[g'(N) = \frac{b(k+N)^2 - bN \cdot 2(k+N)}{((k+N)^2)^2}.\]

Step 5: Simplify the Expression

Simplify the expression in the numerator:\[ b(k+N)^2 - 2bN(k+N) = b[(k+N)^2 - 2N(k+N)] = b(k^2 + 2kN + N^2 - 2kN - 2N^2).\] This simplifies further to:\[ b(k^2 - N^2).\] The denominator remains \((k+N)^4\). Thus, \[ g'(N) = \frac{b(k^2 - N^2)}{(k+N)^4}.\]

Key Concepts

The Quotient Rule in Differentiation

In calculus, the quotient rule is a key differentiation technique used when dealing with the division of two functions. If a function can be represented as a quotient, that is, one function divided by another, then the quotient rule is applied.

For a function expressed as \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of some variable, the derivative is determined using the formula:

• \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]

The rule works by taking the derivative of the numerator \( u \) (notated as \( u' \)) and the derivative of the denominator \( v \) (notated as \( v' \)), then applying them in the formula.

The structure of the quotient rule emphasizes subtracting the product of the original denominator's derivative from the product of the original function's derivative. Finally, this result is divided by the square of the denominator.

In the given exercise, using this rule helps simplify otherwise complex derivatives into a manageable expression. This technique ensures that differentiation is efficient and straightforward.

Understanding the Chain Rule

The chain rule is another essential technique in calculus, used predominantly when differentiating composite functions. It allows us to handle functions nested within other functions, by breaking the differentiation process into more straightforward steps.

When you have a composite function \( f(g(x)) \), the chain rule states that the derivative is:

• \[ (f(g(x)))' = f'(g(x)) \cdot g'(x) \]

This means you first differentiate the outer function leaving the inner function intact, then multiply by the derivative of the inner function itself.

In the exercise, the denominator \( (k+N)^2 \) required the application of the chain rule. Here, \( k+N \) is an inner function, while squaring is the outer function. Differentiating the outer function gives us \( 2(k+N) \), then multiplying by the derivative of the inner gives \( 2(k+N) \cdot 1 \) (since the derivative of \( k+N \) is 1).

The chain rule simplifies the process of handling layers of functions, making it indispensable for dealing with complex function relationships in calculus.

Differentiation: What Does Derivative of a Function Mean?

The derivative of a function represents the rate at which a function's value is changing at any given point, effectively portraying the function's slope at that specific point. In essence, it's a fundamental concept that lets us understand the behavior of functions dynamically.

A derivative is foundational for analyzing a function’s increasing or decreasing behavior over its domain. In calculus, we notate the derivative of a function \( f(x) \) with respect to \( x \) as \( f'(x) \) or \( \frac{df}{dx} \).

Consider a function that outputs a height value for each time input. The derivative of this function would tell us how fast height is changing at any specified time.

• If the derivative is positive, the function is increasing at that point.
• If it's negative, the function is decreasing.
• If the derivative is zero, the function has a horizontal tangent at that point, indicating a potential maximum or minimum.

Through differentiation, we gain the ability to conduct deep analyses of functions, optimizing them, evaluating limits, and solving real-world problems that range from physics to economics.